Are you asking whether the notion of "limit of a diagram (or functor)" in category theory can be literally seen as a special case of "limit of a function (or limit of a sequence, or limit point of a set, or something like that)" in topology?
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Tom GoodwillieApr 27 '11 at 13:59

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It may seem a simple question but that does not mean there is a simple answer!
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Tim PorterApr 27 '11 at 14:49

There is a topology on the set of arrows of a category you can define as follows: a subset F is closed if every arrow whose codomain is the domain of some arrow in F belongs to F. With this topology every functor (regarded as a function between arrows) is continuous, so you actually have a faithful functor from Cat to Top. The limiting cones corresponding to a diagram D: C -> E belong to the closure of the image of D, but so does every cone. I feel though this is as far as you can get if you try to use the usual notion of topology here. So something different is needed.
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godelianApr 27 '11 at 22:56

1 Answer
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As I stated in the comments, the problem is usually that such topologies do not distinguish between cones and universal cones. However, in the special case where the categories are preorders (i.e., categories where each $Hom(A, B)$ has at most one arrow), then it is possible to give a topology on the set of objects in such a way that for complete preorders, a functor is continuous with respect to the topologies considered if and only if it preserves limits (i.e., inf) of totally (pre)ordered diagrams. Let us call a functor between complete preorders weakly continuous if it does exactly that, i.e., if it preserves all limits corresponding to totally (pre)ordered diagrams. We can topologize the preorders as follows:

Consider first the category $\mathbf{2}$ consisting of an arrow $0 \to 1$ between two objects. Define a topology there which has the object $1$ as the non-trivial closed set. For a general preorder $\mathcal{C}$, define now a topology as the one whose closed sets are of the form $F^{-1}(1)$ for weakly continuous functors $F: \mathcal{C} \to \mathbf{2}$. To see that it is indeed a topology, note that if $(A_i)_i$ are closed sets corresponding to the preimages of $1$ of the weakly continuous functors $(F_i)_i$, respectively, then $\prod_i F_i$ is a weakly continuous functor such that the preimage of $1$ is exactly $\cap_i A_i$. Similarly, if $A, B$ are closed sets corresponding to the preimages of $1$ of the weakly continuous functors $F, G$, respectively, then $F \coprod G$ is weakly continuous and the preimage of $1$ is exactly $A \cup B$ (note that in general the coproduct of two limit preserving functors from preorders to $\mathbf{2}$ need not be limit preserving, so we do need to restrict our considerations to weakly continuous functors).

It is clear that if $F: \mathcal{C} \to \mathcal{D}$ is weakly continuous, it is continuous with respect to the topologies defined above, since for a closed set $A$ in $\mathcal{D}$, preimage of $1$ of the functor $G$, we have that $F^{-1}(A)$ is the preimage of $1$ of the weakly continuous composition $GF$. Conversely, let us see that if $F$ is continuous with respect to the topologies then it must be weakly continuous. If $(C \to C_i)_i$ is a limiting cone in $\mathcal{C}$ corresponding to a totally (pre)ordered diagram $(C_i)_i$, then by definition $C$ belongs to the closure of $(C_i)_i$, and hence $F(C)$ must belong to the closure of $(F(C_i))_i$ in $\mathcal{D}$. If $(F(C) \to F(C_i))_i$ were not a universal cone, let $D$ be the vertex of such a cone; we have an induced arrow $F(C) \to D$ (and hence no arrow $D \to F(C)$). But then the representable functor $[D, -]$ (regarded as a functor with values in $\mathbf{2}$) would be weakly continuous (in fact, limit preserving), and the closed set which is the preimage of $1$ contains each $F(C_i)$ but not $F(C)$, which is absurd, since in that case $F(C)$ would not belong to such a closed set containing the $F(C_i)$. Therefore, $(F(C) \to F(C_i))_i$ must be a universal cone and the proof is complete.

I disagree that $F \cup G : C \to 2$ is limit-preserving if $F$ and $G$ are. Let $C$ be $2 \times 2$, and let $F$ and $G$ be the two projections. Clearly, each preserves limits; however, $F \cup G$ sends $(1, 0)$ and $(0, 1)$ both to $1$, while sending their infimum $(0, 0)$ to $0$.
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Sridhar RameshMay 8 '11 at 23:52

@ Sridhar: Indeed you are right. I thought I had a proof but I realized from your counterexample I was implicitely assuming every hom-set had one arrow, i.e., that it is actually a total order! I'll try to check if this can be solved or not.
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godelianMay 9 '11 at 0:25

So far I can only think of a very ad hoc argument to fix this: what if instead of a topology we define a "pseudotopology" by relaxing a bit the usual definition? The intersection of closed subsets of the preorder is closed, but the union of two such subsets is closed only when such a union is a total (pre)order, i.e., each hom-set in the union has exactly one arrow. The rest of the argument follows. But the question is, is that notion of pseudotopology useful?
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godelianMay 9 '11 at 1:41

I have found a better way to fix this, although the result is now weaker. I have edited accordingly.
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godelianMay 9 '11 at 3:28